Table Of ContentAdvanced Modern
Engineering Mathematics
Glyn James
fourth edition
Solutions Manual
Solutions Manual
Advanced Modern
Engineering Mathematics
4
th edition
Glyn James
ISBN 978-0-273-71925-0
c⃝Pearson Education Limited 2011
Lecturers adopting the main text are permitted to download the manual as
required.
c⃝Pearson Education Limited 2011
i
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This edition published 2011
c⃝Pearson Education Limited 2011
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ISBN: 978-0-273-71925-0
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ii
TABLE OF CONTENTS
Page
Chapter 1. Matrix Analysis
1
Chapter 2. Numerical Solution of Ordinary Differential Equations
86
Chapter 3. Vector Calculus
126
Chapter 4. Functions of a Complex Variable
194
Chapter 5. Laplace Transforms
270
Chapter 6. The z Transform
369
Chapter 7. Fourier Series
413
Chapter 8. The Fourier Transform
489
Chapter 9. Partial Differential Equations
512
Chapter 10. Optimization
573
Chapter 11. Applied Probability and Statistics
639
iii
1
Matrix Analysis
Exercises 1.3.3
1(a)
Yes,
as the three vectors are linearly independent and span three-
dimensional space.
1(b)
No, since they are linearly dependent
⎡
⎣
3
2
5
⎤
⎦ − 2
⎡
⎣
1
0
1
⎤
⎦ =
⎡
⎣
1
2
3
⎤
⎦
1(c)
No, do not span three-dimensional space.
Note, they are also linearly
dependent.
2
Transformation matrix is
A =
1
√
2
⎡
⎣
1
1
0
1
−1
0
0
0
√
2
⎤
⎦
⎡
⎣
1
0
0
0
1
0
0
0
1
⎤
⎦ =
⎡
⎣
1
√
2
1
√
2
0
1
√
2
− 1
√
2
0
0
0
1
⎤
⎦
Rotates the (e1, e2) plane through π/4 radians about the e3 axis.
3
By checking axioms (a)–(h) on p. 10 it is readily shown that all cubics
ax3 + bx2 + cx + d form a vector space. Note that the space is four dimensional.
3(a)
All cubics can be written in the form
ax3 + bx2 + cx + d
and {1, x, x2, x3} are a linearly independent set spanning four-dimensional space.
Thus, it is an appropriate basis.
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Glyn James, Advanced Modern Engineering Mathematics, 4th Edition
3(b)
No, does not span the required four-dimensional space. Thus a general
cubic cannot be written as a linear combination of
(1 − x), (1 + x), (1 − x3), (1 + x3)
as no term in x2 is present.
3(c)
Yes as linearly independent set spanning the four-dimensional space
a(1 − x) + b(1 + x) + c(x2 − x3) + d(x2 + x3)
= (a + b) + (b − a)x + (c + a)x2 + (d − c)x3
≡ α + βx + γx2 + δx3
3(d)
Yes as a linear independent set spanning the four-dimensional space
a(x − x2) + b(x + x2) + c(1 − x3) + d(1 + x3)
= (a + b) + (b − a)x + (c + d)x2 + (d − c)x3
≡ α + βx + γx2 + δx3
3(e)
No not linearly independent set as
(4x3 + 1) = (3x2 + 4x3) − (3x2 + 2x) + (1 + 2x)
4
x + 2x3, 2x − 3x5, x + x3 form a linearly independent set and form a basis
for all polynomials of the form α + βx3 + γx5. Thus, S is the space of all odd
quadratic polynomials. It has dimension 3.
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3
Exercises 1.4.3
5(a)
Characteristic polynomial is λ3 − p1λ2 − p2λ − p3 with
p1 = trace A = 12
B1 = A − 12I =
⎡
⎣
−9
2
1
4
−7
−1
2
3
−8
⎤
⎦
A2 = A B1 =
⎡
⎣
−17
−5
−7
−18
−30
7
2
−5
−33
⎤
⎦
p2 = 1
2 trace A2 = −40
B2 = A2 + 40I =
⎡
⎣
23
−5
−7
−18
10
7
2
−5
7
⎤
⎦
A3 = A B2 =
⎡
⎣
35
0
0
0
35
0
0
0
35
⎤
⎦
p3 = 1
3 trace A3 = 35
Thus, characteristic polynomial is
λ3 − 12λ2 + 40λ − 35
Note that B3 = A3 − 35I = 0 confirming check.
5(b)
Characteristic polynomial is λ4 − p1λ3 − p2λ2 − p3λ − p4 with
p1 = trace A = 4
B1 = A − 4I =
⎡
⎢⎣
−2
−1
1
2
0
−3
1
0
−1
1
−3
1
1
1
1
−4
⎤
⎥⎦
A2 = A B1 =
⎡
⎢⎣
−3
4
0
−3
−1
−2
−2
1
2
0
−2
−5
−3
−3
−1
3
⎤
⎥⎦
⇒ p2 = 1
2 trace A2 = −2
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Glyn James, Advanced Modern Engineering Mathematics, 4th Edition
B2 = A2 + 2I =
⎡
⎢⎣
−1
4
0
−3
−1
0
−2
1
2
0
0
−5
−3
−3
−1
5
⎤
⎥⎦
A3 = A B2 =
⎡
⎢⎣
−5
2
0
−2
1
0
−2
−4
−1
−7
−3
4
0
4
−2
−7
⎤
⎥⎦
⇒ p3 = 1
3 trace A3 = −5
B3 = A3 + 5I =
⎡
⎢⎣
0
0
0
−2
1
5
−2
−4
−1
−8
2
4
0
4
−2
−2
⎤
⎥⎦
A4 = A B3 =
⎡
⎢⎣
−2
0
0
0
0
−2
0
0
0
0
−2
0
0
0
0
−2
⎤
⎥⎦
⇒ p4 = 1
4trace A4 = −2
Thus, characteristic polynomial is λ4 − 4λ3 + 2λ2 + 5λ + 2
Note that B4 = A4 + 2I = 0 as required by check.
6(a)
Eigenvalues given by
�� 1−λ
1
1
1−λ
�� = λ2 − 2λ = λ(λ − 2) = 0
so eigenvectors are λ1 = 2, λ2 = 0
Eigenvectors given by corresponding solutions of
(1 − λi)ei1 + ei2 = 0
ei1 + (1 − λi)ei2 = 0
Taking i = 1, 2 gives the eigenvectors as
e1 = [1 1]T , e2 = [1 − 1]T
(1)
6(b)
Eigenvalues given by
�� 1−λ
3
2
2−λ
�� = λ2 − 3λ − 4 = (λ + 1)(λ − 4) = 0
so eigenvectors are λ1 = 4, λ2 = −1
Eigenvectors given by corresponding solutions of
(l − λi)ei1 + 2ei2 = 0
3ei1 + (2 − λi)ei2 = 0
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Taking i = 1, 2 gives the eigenvectors as
e1 = [2 3]T , e2 = [1 − 1]T
6(c)
Eigenvalues given by
������
1 − λ
0
−4
0
5 − λ
4
−4
4
3 − λ
������
= λ3 + 9λ2 + 9λ − 81 = (λ − 9)(λ − 3)(λ + 3) = 0
So the eigenvalues are λ1 = 9, λ2 = 3, λ3 = −3.
The eigenvectors are given by the corresponding solutions of
(1 − λi)ei1 + 0ei2 − 4ei3 = 0
0ei1 + (5 − λi)ei2 + 4ei3 = 0
−4ei1 + 4ei2 + (3 − λi)ei3 = 0
Taking i = 1, λi = 9 solution is
e11
8
= −e12
16 = e13
−16 = β1
⇒ e1 = [−1 2 2]T
Taking i = 2, λi = 3 solution is
e21
−16 = −e22
16 = e23
8
= β2
⇒ e2 = [2 2 − 1]T
Taking i = 3, λi = −3 solution is
e31
32 = −e32
16 = e33
32 = β3
⇒ e3 = [2 − 1 2]T
6(d)
Eigenvalues given by
������
1 − λ
1
2
0
2 − λ
2
−1
1
3 − λ
������
= 0
Adding column 1 to column 2 gives
������
1 − λ
2 − λ
2
0
2 − λ
2
−1
0
3 − λ
������
= (2 − λ)
������
1 − λ
1
2
0
1
2
−1
0
3 − λ
������
R1−R2(2 − λ)
������
1 − λ
0
0
0
1
2
−1
0
3 − λ
������
= (2 − λ)(1 − λ)(3 − λ)
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Glyn James, Advanced Modern Engineering Mathematics, 4th Edition
so the eigenvalues are λ1 = 3, λ2 = 2, λ3 = 1.
Eigenvectors are the corresponding solutions of (A − λiI)ei = 0
When λ = λ1 = 3 we have
⎡
⎣
−2
1
2
0
−1
2
−1
1
0
⎤
⎦
⎡
⎣
e11
e12
e13
⎤
⎦ = 0
leading to the solution
e11
−2 = −e12
2
= e13
−1 = β1
so the eigenvector corresponding to λ2 = 3 is e1 = β1[2 2 1]T , β1 constant.
When λ = λ2 = 2 we have
⎡
⎣
−1
1
2
0
0
2
−1
1
1
⎤
⎦
⎡
⎣
e21
e22
e23
⎤
⎦ = 0
leading to the solution
e21
−2 = −e22
2
= e23
0 = β3
so the eigenvector corresponding to λ2 = 2 is e2 = β2[1 1 0]T , β2 constant.
When λ = λ3 = 1 we have
⎡
⎣
0
1
2
0
1
2
−1
1
2
⎤
⎦
⎡
⎣
e31
e32
e33
⎤
⎦ = 0
leading to the solution
e31
0
= −e32
2
= e33
1 = β1
so the eigenvector corresponding to λ3 = 1 is e3 = β3[0 − 2 1]T , β3 constant.
6(e)
Eigenvalues given by
������
5 − λ
0
6
0
11 − λ
6
6
6
−2 − λ
������
= λ3 − 14λ2 − 23λ − 686 = (λ − 14)(λ − 7)(λ + 7) = 0
so eigenvalues are λ1 = 14, λ2 = 7, λ3 = −7
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Eigenvectors are given by the corresponding solutions of
(5 − λi)ei1 + 0ei2 + 6ei3 = 0
0ei1 + (11 − λi)ei2 + 6ei3 = 0
6ei1 + 6ei2 + (−2 − λi)ei3 = 0
When i = 1, λ1 = 14 solution is
e11
12 = −e12
−36 = e13
18 = β1 ⇒ e1 = [2 6 3]T
When i = 2, λ2 = 7 solution is
e21
−72 = −e22
−36 = e23
−24 = β2 ⇒ e2 = [6 − 3 2]T
When i = 3, λ3 = −7 solution is
e31
54 = −e32
−36 =
e33
−108 = β3 ⇒ e3 = [3 2 − 6]T
6(f)
Eigenvalues given by
������
1 − λ
−1
0
1
2 − λ
1
−2
1
−1 − λ
������
R1+R2
������
−1 − λ
0
−1 − λ
1
2 − λ
1
−2
1
−1 − λ
������
= (1 + λ)
������
−1
0
0
1
2 − λ
0
−2
1
1 − λ
������
= 0, i.e. (1 + λ)(2 − λ)(1 − λ) = 0
so eigenvalues are λ1 = 2, λ2 = 1, λ3 = −1
Eigenvectors are given by the corresponding solutions of
(1 − λi)ei1 − ei2 + 0ei3 = 0
ei1 + (2 − λi)ei2 + ei3 = 0
−2ei1 + ei2 − (1 + λi)ei3 = 0
Taking i = 1, 2, 3 gives the eigenvectors as
e1 = [−1 1 1]T , e2 = [1 0 − 1]T , e3 = [1 2 − 7]T
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Glyn James, Advanced Modern Engineering Mathematics, 4th Edition
6(g)
Eigenvalues given by
������
4 − λ
1
1
2
5 − λ
4
−1
−1
−λ
������
R1 + (R2 + R3)
������
5 − λ
5 − λ
5 − λ
2
5 − λ
4
−1
−1
−λ
������
= (5 − λ)
������
1
0
0
2
3 − λ
2
−1
0
1 − λ
������
= (5 − λ)(3 − λ)(1 − λ) = 0
so eigenvalues are λ1 = 5, λ2 = 3, λ3 = 1
Eigenvectors are given by the corresponding solutions of
(4 − λi)ei1 + ei2 + ei3 = 0
2ei1 + (5 − λi)ei2 + 4ei3 = 0
−ei1 − ei2 − λiei3 = 0
Taking i = 1, 2, 3 and solving gives the eigenvectors as
e1 = [2 3 − 1]T , e2 = [1 − 1 0]T , e3 = [0 − 1 1]T
6(h)
Eigenvalues given by
������
1 − λ
−4
−2
0
3 − λ
1
1
2
4 − λ
������
R1+2R2
������
1 − λ
2 − 2λ
0
0
3 − λ
1
1
2
4 − λ
������
= (1 − λ)
������
1
0
0
0
3 − λ
1
1
0
4 − λ
������
= (1 − λ)(3 − λ)(4 − λ) = 0
so eigenvalues are λ1 = 4, λ2 = 3, λ3 = 1
Eigenvectors are given by the corresponding solutions of
(1 − λi)ei1 − 4ei2 − 2ei3 = 0
2ei1 + (3 − λi)ei2 + ei3 = 0
ei1 + 2ei2 + (4 − λi)ei3 = 0
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Taking i = 1, 2, 3 and solving gives the eigenvectors as
e1 = [2 − 1 − 1]T , e2 = [2 − 1 0]T , e3 = [4 − 1 − 2]T
Exercises 1.4.5
7(a)
Eigenvalues given by
������
2 − λ
2
1
1
3 − λ
1
1
2
2 − λ
������
R1−R2
������
1 − λ
−1 + λ
0
0
3 − λ
1
1
2
2 − λ
������
= (1 − λ)
������
1
0
0
1
4 − λ
1
1
3
2 − λ
������
= (1 − λ)[λ2 − 6λ + 5] = (1 − λ)(λ − 1)(λ − 5) = 0
so eigenvalues are λ1 = 5, λ2 = λ3 = 1
The eigenvectors are the corresponding solutions of
(2 − λi)ei1 + 2ei2 + ei3 = 0
ei1 + (3 − λi)ei2 + ei3 = 0
ei1 + 2ei2 + (2 − λi)ei3 = 0
When i = 1, λ1 = 5 and solution is
e11
4
= −e12
−4 = e13
4
= β1 ⇒ e1 = [1 1 1]T
When λ2 = λ3 = 1 solution is given by the single equation
e21 + 2e22 + e23 = 0
Following the procedure of Example 1.6 we can obtain two linearly independent
solutions. A possible pair are
e2 = [0 1 2]T , e3 = [1 0 − 1]T
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Glyn James, Advanced Modern Engineering Mathematics, 4th Edition
7(b)
Eigenvalues given by
������
−λ
−2
−2
−1
1 − λ
2
−1
−1
2 − λ
������
= −λ3 + 3λ2 − 4 = −(λ + 1)(λ − 2)2 = 0
so eigenvalues are λ1 = λ2 = 2, λ3 = −1
The eigenvectors are the corresponding solutions of
−λiei1 − 2ei2 − 2ei3 = 0
−ei1 + (1 − λi)ei2 + 2ei3 = 0
−ei1 − ei2 + (2 − λi)ei3 = 0
When i = 3, λ3 = −1 corresponding solution is
e31
8
= −e32
−1 = e33
3
= β3 ⇒ e3 = [8 1 3]T
When λ1 = λ2 = 2 solution is given by
−2e21 − 2e22 − 2e23 = 0
(1)
−e21 − e22 + 2e23 = 0
(2)
−e21 − e22 = 0
(3)
From (1) and (2) e23 = 0 and it follows from (3) that e21 = −e22 . We deduce that
there is only one linearly independent eigenvector corresponding to the repeated
eigenvalues λ = 2. A possible eigenvector is
e2 = [1 − 1 0]T
7(c)
Eigenvalues given by
������
4 − λ
6
6
1
3 − λ
2
−1
−5
−2 − λ
������
R1−3R3
������
1 − λ
−3 + 3λ
0
1
3 − λ
2
−1
−5
−2 − λ
������
= (1 − λ)
������
1
−3
0
1
3 − λ
2
−1
−5
−2 − λ
������
= (1 − λ)
������
1
0
0
1
6 − λ
2
1
−8
−2 − λ
������
= (1 − λ)(λ2 + λ + 4) = (1 − λ)(λ − 2)2 = 0
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so eigenvalues are λ1 = λ2 = 2, λ3 = 1.
The eigenvectors are the corresponding solutions of
(4 − λi)ei1 + 6ei2 + 6ei3 = 0
ei1 + (3 − λi)ei2 + 2ei3 = 0
−ei1 − 5ei2 − (2 + λi)ei3 = 0
When i = 3, λ3 = 1 corresponding solution is
e31
4
= −e32
−1
= e33
−3 = β3 ⇒ e3 = [4 1 − 3]T
When λ1 = λ2 = 2 solution is given by
2e21 + 6e22 + 6e23 = 0
e21 + e22 + 2e23 = 0
−e21 − 5e22 − 4e23 = 0
so that
e21
6
= −e22
−2
= e23
−4 = β2
leading to only one linearly eigenvector corresponding to the eigenvector λ = 2. A
possible eigenvector is
e2 = [3 1 − 2]T
7(d)
Eigenvalues given by
������
7 − λ
−2
−4
3
−λ
−2
6
−2
−3 − λ
������
R1−2R2
������
1 − λ
−2 + 2λ
0
3
−λ
−2
6
−2
−3 − λ
������
= (1 − λ)
������
1
−2
0
3
−λ
−2
6
−2
−3 − λ
������
= (1 − λ)
������
1
0
0
3
6 − λ
−2
6
10
−3 − λ
������
= (1 − λ)(λ − 2)(λ − 1) = 0
so eigenvalues are λ1 = 2, λ2 = λ3 = 1.
The eigenvectors are the corresponding solutions of
(7 − λi)ei1 − 2ei2 − 4ei3 = 0
3ei1 − λiei2 − 2ei3 = 0
6ei1 − 2ei2 − (3 + λi)ei3 = 0
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